Question: Solve for $x$ : $ 8|x + 3| - 2 = -4|x + 3| + 8 $
Explanation: Add $ {4|x + 3|} $ to both sides: $ \begin{eqnarray} 8|x + 3| - 2 &=& -4|x + 3| + 8 \\ \\ { + 4|x + 3|} && { + 4|x + 3|} \\ \\ 12|x + 3| - 2 &=& 8 \end{eqnarray} $ Add ${2}$ to both sides: $ \begin{eqnarray} 12|x + 3| - 2 &=& 8 \\ \\ { + 2} &=& { + 2} \\ \\ 12|x + 3| &=& 10 \end{eqnarray} $ Divide both sides by ${12}$ $ \dfrac{12|x + 3|} {{12}} = \dfrac{10} {{12}} $ Simplify: $ |x + 3| = \dfrac{5}{6}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 3 = -\dfrac{5}{6} $ or $ x + 3 = \dfrac{5}{6} $ Solve for the solution where $x + 3$ is negative: $ x + 3 = -\dfrac{5}{6} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& -\dfrac{5}{6} \\ \\ {- 3} && {- 3} \\ \\ x &=& -\dfrac{5}{6} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $6$ $ x = - \dfrac{5}{6} {- \dfrac{18}{6}} $ $ x = -\dfrac{23}{6} $ Then calculate the solution where $x + 3$ is positive: $ x + 3 = \dfrac{5}{6} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& \dfrac{5}{6} \\ \\ {- 3} && {- 3} \\ \\ x &=& \dfrac{5}{6} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $6$ $ x = \dfrac{5}{6} {- \dfrac{18}{6}} $ $ x = -\dfrac{13}{6} $ Thus, the correct answer is $x = -\dfrac{23}{6} $ or $x = -\dfrac{13}{6} $.